See A322322 for the definition of a stuffable number in general.
The sequence is infinite because it contains the numbers 202020...20 (that is, k copies of 20) for any odd k.  Hans Havermann, Dec 03 2018
Do we have enough terms now to make a guess about exactly what numbers are in this sequence?  N. J. A. Sloane, Dec 06 2018
From M. F. Hasler, Dec 07 2018: (Start)
Def. 1: Let E(n) = sumdigits(n)  #digits(n) = A007953(n)  A055642(n). We call N > 0 admissible iff E(N) = A010879(N), the last digit of N.
Remark: Any term of this sequence must be admissible. If N is admissible then any permutation of N's digits with fixed last digit (and no leading 0) is again admissible. Also, the last digit of N can be replaced by any arbitrary digit.
Prop. 1: If E(n) >= 0, then N = n*10^E(n) is admissible. (Obvious from definitions.)
Remark: Most terms of this sequence are of that form, with some n not a multiple of 10. Are there any terms not of this form, beyond a(2) = 22 and a(4) = 126?
Prop. 2: If N is admissible and a multiple of 10, then the concatenation of N and any other admissible number M is again admissible. In particular, M can be 1 or any number of concatenations of N. Instead of N one can take an arbitrary number of admissible multiples of 10, (N1, ..., Nk).
Def. 2: When N is admissible, let S(N) be the result of N stuffed with itself, i.e., after each nonzero digit d[i] of N except for the last digit, we insert d[i] digits of N, taken one after the other starting with d[1], d[2], etc.: after d[1] are inserted d[1], ..., d[d[1]]; after d[2], if nonzero, are inserted d[d[1]+1], ..., d[d[1]+d[2]], etc.
Remark: The last digit inserted, just before the last digit d[L], is also d[d[1]+...+d[L1]] = d[L]. Thus, the result S(N) has both its first and its last digit occurring twice in a row.
Def. 3: We say N is selfstuffable iff N is admissible and N divides S(N).
Example: For 0 < k < 9, N = (k+1)*10^k is selfstuffable, thus in this sequence.
Prop. 3: If N and M are admissible and N = 0 (mod 10), then S(concat(N,M)) = concat(S(N),S(M)). The same is true with any number of copies of N.
Lemma: If N is admissible, then S(N*10^E(N)) = S(N)*10^E(N). (Similar to Prop. 1.)
Corollary: If M is admissible, then so are N = M*10^E(M) and T = concat(N,N,M) = (100^L + 10^L + 1)*M with L = #digits(N), and S(T) = (10^(4L) + 100^L + 1)*S(M) = (100^L  10^L + 1)*T*S(M)/M. Thus, if N is selfstuffable, then so is T.
More generally, Sum_{k=0..m} x^(2k) = Sum_{k=0..m} (x)^k * Sum_{k=0..m} x^k for any even m, and therefore (with x = 10^L):
Theorem: If M is selfstuffable, then so is M_m := Sum_{k=0..2m} 10^(k*L)*M with L = #digits(M) + E(M) for all m >= 0. (This is the concatenation of 2m copies of M*10^E(M) and then M.)
Example: For M = 20, E(M) = 0, this is Hans Havermann's comment above. For M = 22, E(M) = 2, it asserts that 22002200...22 (any odd number of '22's) is in the sequence; similarly for M = 126, E(M) = 6, not only 126*10^6 but also 126*(10^18 + 10^9 + 1) etc. are selfstuffable.
We may call a selfstuffable (or an admissible) number primitive if it is not of the form M_m with m > 0. The smallest integer M > 0 such that a(n) = M_m or a(n) = (M*10^E(M))_m will be called the root of a(n). It is easy to see that any term has a root which is not a multiple of 10.
The sequence can be computed very efficiently as follows: Consider numbers M > 0 that are not a multiple of 10. If E(M) > 0 and N  S(N) for N = M*10^E(M), then all {N_m; m>=0} are in the sequence. If in addition to this M is admissible and M  S(M), then all {M_m; m>=0} are also in the sequence. This yields all terms. (End)
The list of roots is now given in A322002.  M. F. Hasler, Dec 09 2018
For nonnegative k, 10^9*(103*10^k+(10^k1)/9)+8*10^6 will always be selfstuffable, thus ensuring that there will always be at least one ddigit solution when d > 11.  Hans Havermann, Dec 12 2018
